Find the value of b for which
(summation function; n=0, infinity) e^nb = 1 + e^b + e^2b + e^3b + ··· = 9
How do I go about doing this? Thanks!
Hi bhaktir,
$\displaystyle \sum_{n=0}^{\infty}e^{nb}=1 + e^{b} + e^{2b} + e^{3b} +\cdots= 9$
$\displaystyle e^{b} + e^{2b} + e^{3b} + \cdots= 8$
$\displaystyle e^{b}\left(1 + e^{b} + e^{2b} + e^{3b} +\cdots \right)=8$
$\displaystyle e^{b}\sum_{n=0}^{\infty}e^{nb}=8$
$\displaystyle 9e^{b}=8~;~\mbox{Since, }\sum_{n=0}^{\infty}e^{nb}=9$
$\displaystyle e^{b}=\frac{8}{9}$
$\displaystyle b=\ln\left(\frac{8}{9}\right)$